Question

An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will she need to be 95% confident that her sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that σ = 40 seconds.

Answer 1

Answer:

So, the required sample size is $28.$, that she will need to be 95% confident that her sample mean will be within 15 seconds of the true mean.

Step-by-step explanation:

Given that,

Standard deviation ($\sigma$) = 40

and we have to find how large a sample is needed to be 95% confident that her sample mean will be within 15 seconds of the true mean.

Now,

If $x bar$ is used as an estimate of $\mu$, we can be $100\times (1-\alpha)$% confident that the error will not exceed a specified amount $e$ when the sample size is

          $n= [\frac{Z_{\frac{\alpha}{2} } ^{}\times \sigma }{e}]^{2}$    ........ (i)

                              where $Z_{\frac{\alpha}{2} }$ is the z value leaving an area of $\frac{\alpha}{2}$ to the right.

So,

      $(100) \times (1-\alpha)$% = 95%

      $1-\alpha=0.95$

            $\alpha = 0.05$

            $\frac{\alpha}{2} = 0.025$

We have $\sigma = 40,$  $\frac{\alpha}{2} = 0.025,$  and $e = 15$, Now using equation (i), the required sample size is,

       $n = [\frac{Z_{0.025}\times 40 }{15}]^{2}$      ........(ii)

Now we have to find the value of $Z_{0.025}$.

So, the $Z_{\frac{\alpha}{2} } = Z_{0.025}$ is the z-value leaving an area of 0.025 to the right {the area left of the $Z_{0.025}$ is (1-0.025) = 0.975.}

Using Normal Probability table, we see that the closest z-value which leaving an area of 0.025 to the right (i.e. an area of 0.975 to the left) is $Z_{0.025} = 1.96$

Now,

Using equation (ii),

                               $n= [\frac{Z_{0.025}\times 40 }{15}]^{2}$

                                   = $[\frac{1.96\times 40}{15}]^{2}$

                                   ≈ $27.3$

So, the required sample size is $28.$